Question #e32b6

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Question #e32b6

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Answer 1 · 2015-09-15T18:03:44

$I_{3 (a q)}^{-} + 2 S_{2} O_{3 (a q)}^{2 -} \to 3 I_{(a q)}^{-} + S_{4} O_{6 (a q)}^{2 -}$ $I_(3(aq))^(-) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)$

Explanation:

Oxidation:

$2 S_{2} O_{3 (a q)}^{2 -} \to S_{4} O_{6 (a q)}^{2 -} + 2 e^{-}$ $2S_2O_(3(aq))^(2–) -> S_4O_(6(aq))^(2–) + 2e^-$

Reduction: $I_{3 (a q)}^{-} + 2 e^{-} \to 3 I_{(a q)}^{-}$ $I_(3(aq))^(–) + 2e^(-) -> 3 I_((aq))^(–)$

the overall reaction then:

Redox:

$I_{3 (a q)}^{-} + 2 S_{2} O_{3 (a q)}^{2 -} \to 3 I_{(a q)}^{-} + S_{4} O_{6 (a q)}^{2 -}$ $I_(3(aq))^(–) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)$

Stoichiometry:

From the balanced equation, we can say that:

$\frac{1}{1} \cdot n_{I_{3}^{-}} = \frac{1}{2} \cdot n_{S_{2} O_{3}^{2 -}}$ $(1)/"1" * n_(I_3^(–)) = (1)/"2" * n_(S_2O_3^(2–))$ .

where the denominators 1 and 2 were taking from the corresponding coefficients of $I_{3}^{-}$ $I_3^–$ and $S_{2} O_{3}^{2 -}$ $S_2O_3^(2–)$ respectively.

Now,

$[I_{3}^{-}]$ $[I_3^–]$ . $V = \frac{1}{2} \cdot [S_{2} O_{3}^{2 -}] \cdot V^{'}$ $V = (1)/"2" * [S_2O_3^(2–)] * V^'$

(since $n = C \cdot V$ $n = C * V$ ), where

$V^{'} = 10.50 mL$ $V^' = "10.50 mL "$ and $V = 15.00 mL$ $" "V = "15.00 mL"$

Therefore,

$[I_{3}^{-}] = \frac{1}{2} \cdot [S_{2} O_{3}^{2 -}] \cdot \frac{V'}{V}$ $[I_3^–] = (1)/"2" * [S_2O_3^(2–)] * (V')/"V"$

$[I_{3}^{-}] = \frac{1}{2} \cdot 0.0500M \cdot \frac{10.50 mL}{15.00 mL} = 0.0175M$ $[I_3^–] = (1)/"2" * "0.0500M" * (10.50 cancel("mL"))/(15.00 cancel("mL")) = "0.0175M"$

(rounded to 3 significant figures)

I would like to suggest the following video, that deals with a similar example in terms of stoichiometry of a redox reaction.

Answer 2 · 2015-09-15T21:28:59

$(a) I_{3 (a q)}^{-} + 2 S_{2} O_{3 (a q)}^{2 -} \to S_{4} O_{6 (a q)}^{2 -} + 3 I_{(a q)}^{-}$ $(a)" "I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-$

$(b) 0.0175 mol/l$ $(b)" "0.0175"mol/l"$

Explanation:

Thiosulfate ions give out electrons:

$2 S_{2} O_{3 (a q)}^{2 -} \to S_{4} O_{6 (a q)}^{2 -} + 2 e$ $2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+2e$

These are taken in by $I_{3}^{-}$ $I_3^-$ which are reduced:

$I_{3 (a q)}^{-} + 2 e \to 3 I_{(a q)}^{-}$ $I_(3(aq))^(-)+2erarr3I_((aq))^(-)$

Adding both sides gives:

$I_{3 (a q)}^{-} + 2 S_{2} O_{3 (a q)}^{2 -} \to S_{4} O_{6 (a q)}^{2 -} + 3 I_{(a q)}^{-}$ $I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-$

(b).

$c = \frac{n}{v}$ $c=n/v$

$n = c \times v$ $n=cxxv$

So no. moles $S_{2} O_{3}^{2 -} = 0.05 \times 0.0105 = 5.25 \times 10^{- 4}$ $S_2O_3^(2-)=0.05xx0.0105=5.25xx10^(-4)$

From the equation we can see that the number of moles of $I_{3}^{-}$ $I_3^-$ must be half of this:

no. moles $I_{3}^{-} = \frac{5.25 \times 10^{- 4}}{2} = 2.625 \times 10^{- 4}$ $I_3^(-)=(5.25xx10^(-4))/(2)=2.625xx10^(-4)$

$c = \frac{n}{v}$ $c=n/v$

So $[I_{3 (a q)}^{-}] = \frac{2.625 \times 10^{- 4}}{0.015} = 0.0175 mol/l$ $[I_(3(aq))^(-)]=(2.625xx10^(-4))/(0.015)=0.0175"mol/l"$

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Question #e32b6

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